Go Find It!

I will write a different solution than the solution by @Mark Hennings .

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Note that $x \sin A + y \sin B + z \sin C = \text{Im} (xe^{iA}+ye^{iB}+ ze^{iC})=0$ and let $xe^{iA}=\alpha,ye^{iB}=\beta,ze^{iC}=\gamma$ , where $\alpha + \beta +\gamma \in \mathbb{R}$

Extending the same logic for the second equation , we get $\alpha^2+\beta^2+\gamma^2 \in \mathbb{R}$ which implies $\alpha \beta + \beta \gamma +\gamma \alpha \in \mathbb{R}$ from the identity $(\alpha + \beta + \gamma)^2=\alpha^2+\beta^2+\gamma^2+2(\alpha \beta + \beta \gamma +\gamma \alpha)$

Now consider the identity $\alpha^3+\beta^3+\gamma^3=3\alpha \beta \gamma + (\alpha+\beta+\gamma)(\alpha^2+\beta^2+\gamma^2-(\alpha \beta+\beta \gamma + \gamma \alpha))$

Here $\alpha \beta \gamma=xyze^{i(A+B+C)}=xyze^{i\pi}=-xyz \in \mathbb{R}$ which implies $\alpha^3+\beta^3+\gamma^3 \in \mathbb{R}$

Which means $\text{Im}(x^3e^{i3A}+y^3e^{i3B}+z^3e^{i3C})=0 \implies x^3 \sin 3A + y^3 \sin 3B + z^3 \sin 3C=\boxed{0}$

Suppose that none of $A,B,C$ is a multiple of $\pi$ , so that $\sin A, \sin B, \sin C \neq 0$ . Then $z\sin C = -(x\sin A + y\sin B)$ , and hence $\begin{aligned} 0 & = \; \sin^2C(x^2\sin2A + y^2\sin2B + z^2\sin2C) \; = \; \sin^2C(x^2\sin2A + y^2\sin2B) + (x\sin A + y\sin B)^2\sin2C \\ & = \; x^2(\sin2A\sin^2C + \sin^2A\sin2C) + 2xy\sin A \sin B \sin 2C + y^2(\sin2B\sin^2C + \sin^2B\sin2C) \\ & = \; 2x^2\sin A \sin C(\cos A \sin C + \sin A \cos C) + 2xy\sin A \sin B \sin2C + 2y^2\sin B \sin C(\cos B \sin C + \sin B \cos C) \\ & = \; 2x^2\sin A \sin C\sin(A+C) + 2xy\sin A \sin B \sin 2C + 2y^2\sin B \sin C \sin(B+C) \; = \; 2\sin A \sin B \sin C(x^2 + 2xy\cos C + y^2) \\ & = \; 2\sin A \sin B \sin C\big[(x + y\cos C)^2 + y^2\sin^2C\big] \end{aligned}$ so that $x=y=0$ . The first condition now implies that $z\sin C = 0$ , and hence $x=y=z=0$ , so that $x^3\sin3A + y^3\sin3B + z^3\sin3C = 0$

Suppose now that $C =n\pi$ is a multiple of $\pi$ . Then $A+B = (1-n)\pi$ and so $\sin B = (-1)^{n-1}\sin A$ , $\sin2B = -\sin 2A$ , so that $\sin A(x + (-1)^{n-1}y) \; =\; \sin2A(x^2 - y^2) \; = \; 0$

• If $A$ is not a multiple of $\pi$ we deduce that $x = (-1)^ny$ . But then $x^3 = (-1)^ny^3$ and $\sin3B = (-1)^{n-1}\sin3A$ , while $\sin3C = 0$ , so that $x^3\sin3A + y^3\sin3B + z^3\sin3C = 0$ .
• If $A$ is a multple of $\pi$ , then $B$ is a multiple of $\pi$ as well. Then $x^j\sin jA + y^j\sin jB + z^j \sin jC = 0$ for any $j \in \mathbb{N}$ , whatever the values of $x,y,z$ .

Thus the only possible value of $x^3\sin3A + y^3\sin3B + z^3\sin3C$ is $\boxed{0}$ .